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In the following circuit I want to calculate the current flowing through the voltage source (node A)

schematic

simulate this circuit – Schematic created using CircuitLab

If I use the KCL I get \$I_A = I_{R2} - I_{R1}\$ (I assumed that the current flows always from left to right and from top to bottom). This can be expressed as \$V1({1 \over R2} - {1 \over R1}) = V1( {R2 -R1 \over R1R2})\$

However if I assume that the circuit is equivalent to a voltage source with a parallel resistors \$V_p\$ (obtained from the parallel of \$R1\$ and \$R2\$) the current is \$I_A = {V1 \over R_p} = V1 ( {R1+R2 \over R1R2})\$.

The same result can be obtained with KCL if the current \$I_{R2}\$ and \$I_{R1}\$ are assumed flowing in opposite directions (both entering the node or leaving the node) and also the direction of the current flowing in A makes the difference.

It seem like the voltage source position in the circuit matters. What is the correct answer in this case ? Can I replace the two resistors with an equivalent parallel resistor ?

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  • \$\begingroup\$ Your first expression doesn't make sense for KCL. Instead, you have \$V_1\left(\frac1{R_1}+\frac1{R_2}\right)=I_{_\text{A}}\$. The voltage at the top will induce currents in the two resistors in the same direction relative to the bottom node. Not differing directions. You can assume otherwise. That's fine. But then you need to deal with that fact, appropriately, when you find the sign is different than assumed. \$\endgroup\$ – jonk Jan 12 at 0:14
  • \$\begingroup\$ "Current always flows from left to right and top to bottom" is contradictory for the left mesh. You really, really need to indicate the direction of the currents on your schematic. \$\endgroup\$ – Elliot Alderson Jan 12 at 1:49
  • \$\begingroup\$ @ElliotAlderson I would indicate the current direction if I knew how to do that in CircuitLab. Does the voltage source put a constraint on the current direction ? I mean the resistors are in parallel so the + sign is on the top of the resistors and the current should flows from + to - ( assuming a passive sign convention). If that's the case you can't choose whatever direction you want for the current as stated here: electronics.stackexchange.com/a/301300/174938 \$\endgroup\$ – Bemipefe Jan 13 at 13:22
  • \$\begingroup\$ You can choose whatever direction you like for the assumed direction of current flow. Then you do the analysis. It's possible that a particular current value will turn out to be negative, which is fine. This just means that a positive current flows in the opposite direction. Sometimes it is not possible to infer the direction of every current in the circuit before we start the analysis, so we have to make an assumption in order to set up the equations. \$\endgroup\$ – Elliot Alderson Jan 13 at 14:12
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Each resistor has same voltage over them, so you know how much current flows from power supply to each resistor. And if you get the same result by calculating the combined parallel resistance, then you have proof whether they are parallel or not.

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Your first assumption („current flows from left to right“) is not true and the actual direction in the diagram is meaningless (depending on the convention the current in question just has an inverted sign).

Therefore, I2-I1will actually be 15A-(-30A)=45A while in another convention it would be 15A+30A=45A. Yes, the resistors can be replaced with a parallel resistor of 2 Ohm.

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  • \$\begingroup\$ You can't write a KCL equation unless you know the assumed direction of each current. There is no "convention" for the direction of a current. The currents must be clearly indicated and they absolutely are not meaningless. \$\endgroup\$ – Elliot Alderson Jan 12 at 1:52
  • \$\begingroup\$ that's actually what I was trying to say... you have to make an assumption for each current's direction and that's what I meant with (your personal) "convention". the arrows don't have to or will point in the right direction before you have calculated anything, but in the end they will end up with a negative sign. \$\endgroup\$ – Sim Son Jan 12 at 2:08

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